F1 - Business Maths April 07

BUSINESS MATHEMATICS & QUANTITATIVE METHODS

FORMATION 1 EXAMINATION - APRIL 2007 NOTES

Answer 5 questions. (Only the first 5 questions answered will be marked). All questions carry equal marks.

STATISTICAL FORMULAE TABLES ARE PROVIDED DEPARTMENT OF EDUCATION MATHEMATICS TABLES ARE AVAILABLE ON REQUEST

TIME ALLOWED:

3 hours, plus 10 minutes to read the paper.

INSTRUCTIONS:

During the reading time you may write notes on the examination paper but you may not commence writing in your answer book. Marks for each question are shown. The pass mark required is 50% in total over the whole paper. Start your answer to each question on a new page.

You are reminded that candidates are expected to pay particular attention to their communication skills and care must be taken regarding the format and literacy of the solutions. The marking system will take into account the content of the candidates' answers and the extent to which answers are supported with relevant legislation, case law or examples where appropriate. List on the cover of each answer booklet, in the space provided, the number of each question(s) attempted.

The Institute of Certified Public Accountants in Ireland, 9 Ely Place, Dublin 2.

THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND

BUSINESS MATHEMATICS & QUANTITATIVE METHODS FORMATION 1 EXAMINATION - APRIL 2007 Time Allowed: 3 hours, plus 10 minutes to read the paper

1.

Answer 5 questions Only the first five questions answered will be marked. All questions carry equal marks.

Your company, CPL Products plc is considering two capital projects. These involve the purchase, use and disposal of two production machines. Machine 1 costs €50,000 and Machine 2 costs €45,000. The net cash flows of both machines are set out in the following table. Provide a report to the company on the most economical purchase. Your report should include the Net Present Value and the Internal Rate of Return (IRR) for each machine. The bank will provide an overdraft at a rate of 12% to the company.

Year

Net Cash Flows € Machine 1

Machine 2

1

25,000

13,000

2

24,000

15,000

3

16,000

22,000

4

15,000

35,000

[Total: 20 Marks]

2.

The Equality Officer carried out a survey in 2002 of the income of supervisory and middle management staff in a number of SMEs by their highest educational qualification. The data is set out below. Weekly Income €

Educational Qualification 3rd Level

2nd Level

400 – 500

5

20

500 – 600

6

30

600 – 700

10

25

700 – 800

18

15

800 – 900

21

5

900 – 1000

14

3

1000 – 1100

10

1

1100 - 1200

16

1

You are required to (i)

Calculate the mean and standard deviation for those with 3rd and 2nd level qualifications.

(ii)

Compare the two distributions and comment on your results.

(12 Marks) (8 Marks) [Total: 20 Marks] 1

3.

DIB plc produces specialised glass panels for stores and shops. It is necessary that the panels have sufficient thickness to avoid accidental breakage. The panels are produced with a mean thickness of 4 cms and a standard deviation of 1 cm. The quality control department takes a large number of samples of 100 from the production line. To confirm the quality of the product you are asked to provide the following: (i)

The 95% and 99% confidence intervals and confidence limits for the panels.

(ii)

An explanation, using diagrams, of what these confidence limits mean.

(10 Marks) (10 Marks) [Total: 20 Marks]

4.

The following data was produced by the management accountant to demonstrate the effectiveness of the internal audit function. He claimed that the number of errors discovered in company processes reduced as the frequency of audits increased. Number of errors (y)

Number of audits (x)

19

1

18

2

16

3

16

4

20

5

13

6

6

7

6

8

11

9

9

10

By means of this data: (i)

(ii)

(iii)

Confirm the management accountants claim by calculating the correlation coefficient between the number of errors and the number of audits. (8 Marks) Calculate a linear equation based on the data. (6 Marks) Explain your result and the basis of his claim. (6 Marks) [Total: 20 Marks]

2

5.

As a consultant to CPA Ltd. you are continuing to address a diverse range of business problems. Provide advice to our clients on the following problems.

(i)

A small business wishes to set up a pension plan for its employees. You are aware that personal pensions are calculated on the basis of annual annuities. Calculate how much should be paid as a lump sum now to get an annuity of €20,000 for 6 years at a discount rate of 5%. (8 Marks)

(ii)

The company wishes to purchase a van for the business and is quoted an annual percentage rate (APR) of interest of 20% on a loan of €15,000 over 3 years. Calculate the total interest paid over three years assuming interest is paid at the end of each year. (6 Marks)

(iii)

The leasing company states that the tyres on the van should last for more than 45,000 miles. The tyre production company suggests that the mean life of the tyre is 42,000 miles with a standard deviation of 4,000 miles. Tyre life is normally distributed. Calculate the % of tyres that will last more than 45,000 miles. (6 Marks) [Total: 20 Marks]

6.

You are a member of the management team who has attended a seminar on modern business techniques including the use of business mathematics and quantitative methods. The following terms have been used during presentations at the seminar – ‘Statistical Hypotheses’, ‘The Normal Distribution’, ‘Internal Rate of Return’, ‘The appropriate selection of a Base Period in Index Numbers’. Provide a brief outline on the meaning of these terms for the management team.

[Total: 20 Marks]

END OF PAPER

3

SUGGESTED SOLUTIONS

BUSINESS MATHEMATICS & QUANTITATIVE METHODS FORMATION 1 EXAMINATION - MAY 2007 SOLUTION 1. (i) Net Present Value for each proposal. The scrap values for the machines are included in the ‘Net Cash Flows’ and should not be further considered. A discount value of 20% is included in the table to assist in the calculation of the internal rate of return Cash flows for Machine 1.

Initial cost

Net Cash Flows

50,000

(50,000)

1.000

(50,000)

1.000

(50,000)

1

25,000

0.893

22,325

0.833

20,825

2

24,000

0.797

19,128

0.694

16,656

3

16,000

0.712

11,392

0.579

9,264

4

15,000

0.636

9,540

0.482

7,230

Year

0

PV Factor @ 12%

NPV

PV €

PV Factor @ 20%

12,385

Cash flows for Machine 2.

3,975

4 Marks

Initial cost

Net Cash Flows

45,000

(45,000)

1.000

(45,000)

1.000

(45,000)

1

13,000

0.893

11.609

0.833

10,829

2

15,000

0.797

11,955

0.694

10.410

3

22,000

0.712

15,664

0.579

12,738

4

35,000

0.636

22,260

0.482

16,870

Year

0

NPV

PV Factor @ 12%

PV €

PV €

12,385

PV Factor @ 20%

PV €

5,847

4 Marks

Due to the higher net present value, with this analysis, Machine 2 represents the best proposal for the company. 5

(ii)

Internal rates of return. The IRR can be derived by calculation or graphically. Using the formula N1I2 - N2I1 , where discount rate I1 gives NPV N1 and discount rate I2 gives NPV N2. N1 - N2 Machine 1.

N1 = €12,385, I1 = 12%; N2 = €3,975, I2 = 20%

IRR = 12,385 x 0.2 - 3,975 x 0.12 12,385 - 3,975

Machine 2.

= 2000 = 23.8% 8410

4 Marks

N1 = €16,488, I1 = 12%; N2 = €5,847, I2 = 20%

IRR = 16,488 x 0.2 - 5,847 x 0.12 16,488 - 5,857

= 2,596 = 24.4% 10,641 4 Marks

Summary. NPV €

IRR %

Machine 1

12,385

23.8

Machine 2

16,488

24.4

Machine 2 is the most profitable investment for the company. This is confirmed by both the values of NPV and IRR. Both projects give a return substantially greater than the cost of the overdraft facility provided by the bank 4 Marks [Total: 20 Marks]

6

SOLUTION 2. (i) The mean and standard deviation. To compare the two groups in part (ii) it is necessary to derive the mean and standard deviation for both groups. 3rd level qualifications. Class Interval

f

Mid point x

fx

x - x-

(x – x)2

- 2 f(x – x)

400 – 500

5

450

2,250

- 406

164,836

824,180

500 – 600

6

550

3,300

- 306

93,636

561,816

600 – 700

10

650

6,500

- 206

42,436

424,360

700 – 800

18

750

13,500

- 106

11,236

202,248

800 – 900

21

850

17,850

-

36

756

900 – 1000

14

950

13,300

94

8,836

123,704

1000 – 1100

10

1,050

10,500

194

37,636

376,360

1100 - 1200

16

1,150

18,400

294

86,436

1,382,976

∑

100

6

85,600

Mean = x = ∑ fx ∑f

3,896,400 2 Marks

= 85,600 = 856 100

Standard Deviation. - 2 = √ ∑ f (x - x) ∑f

σ

=

38964

=

197.4

√ 3,896,400

=

100 3 Marks

2nd level qualifications Class Interval

f

Mid point x

fx

x-x

- 2 (x – x)

f(x – x)2

400 – 500

20

450

9,000

- 173

29,929

598,580

500 – 600

30

550

16,500

- 73

5,329

159,870

600 – 700

25

650

16,250

27

729

18,225

700 – 800

15

750

11,250

127

16,129

241,935

800 – 900

5

850

4,250

227

51,529

257,645

900 – 1000

3

950

2,850

327

106,929

320,787

1000 – 1100

1

1,050

1,050

427

183,329

182,329

1100 - 1200

1

1,150

1,150

527

277,729

277,729

∑

100

62,300

7

2,057,100

Mean = x = ∑ fx ∑f

2 Marks

= 62,300 = 623 100

Standard Deviation. σ

=

(ii)

= √ ∑ f (x - x)2 ∑ f

√20,571 =

= √2,057,100 100 3 Marks

143.4

To compare the two groups the most appropriate method is to use the coefficient of variation. This measures the relative dispersion of the two groups. C of V3rd level =

σ/x-

=

197.4/856

=

23.06%

C of V2nd level =

σ/x-

=

143.4/623

=

23.01% 5 Marks

The co-efficient of variation of both distributions is similar. The data with the highest coefficient of variation has the greatest relative dispersion. In this case the relative dispersion between both groups is minimal. Both distributions are skewed showing that a higher number of third level graduates have a greater income. The median would be a more representative average in this particular case. 5 Marks [Total: 20 Marks]

8

SOLUTION 3. i) The general form of a confidence interval is x

=

µ

±

z (σ/√n), where

z1 = 1.96 for the 95% confidence interval z2 = 2.58 for the 99% confidence interval and µ = 4, σ = 1, n = 100.

(ii)

95% confidence interval:

x1 = 4 ± 1.96 (1/√100)

95% confidence limits are:

4.196 and 3.804.

99% confidence interval:

x2 = 4 ± 2.58 (1/√100)

99% confidence limits are:

4.258 and 3.742.

= 4 ± 0.196 5 Marks

= 4 ± 0.258 5 Marks

A confidence interval is a range of values within which a certain level of confidence (95% 0r 99%) can be stated, that is, a range within which a particular value of a variable will lie. The confidence interval for a sample mean is the range of values around the population mean (µ) within which it can be stated, - will lie. normally with 95% or 99% confidence, that a particular sample mean (x) 5 Marks The above problem is represented by the following diagrams.

95% confidence interval. 0.475

0.475

0.025

0.025 X2 3.804

µ=4cm

-2.58

X

X1 4.258

z

+2.58 0

95% confidence interval. 0.495

0.495

0.005

0.005 X2 3.742

µ=4cm

-2.58

X1 4.258 +2.58

0

9

X

z

5 Marks

[Total: 20 Marks]

SOLUTION 4. (i) Correlation co-efficient is derived form the formula:

r =

∑ xy/n - ∑ x/n ∑ y/n √ {∑ x /n - (∑ x/n)2}{∑ y2/n - (∑ y/n)2} 2

Therefore, r =

624/10 - 55/10 x 134/10 √(385/10 - 30.25)(2040/10 - 179.56)

= - 11.3/14.07

(ii)

=

- 0.8

The relationship can be demonstrated by developing a linear regression equation and analysing it. Number of errors Number of audits y x

19 18 16 16 20 13 6 6 11 9 134

1 2 3 4 5 6 7 8 9 10 55

x2

xy

y2

1 4 9 16 25 36 49 64 81 100 385

19 36 48 64 100 78 42 48 99 90 624

361 324 256 256 400 169 36 36 121 81 2040

A linear regression equation may be written as: y = a + bx where ∑ y =

na + b∑ x

∑ xy =

a ∑ x + b ∑ x2 ; where

a

=

∑ y - b∑ x n n

b

=

n∑ xy - ∑ x∑ y, n∑ x2 - (∑ x)2

Inserting values gives b

a

=

10 x 624 - 55 x 134 10 x 385 - 552

=

6240 - 7370 3850 - 3025

=

134 - (-1.37) x 55 10 10

= - 1130 825 =

= - 1.37

13.4 + 7.54 = 20.94 6 Marks

Therefore, y = 20.94 - 1.37x.

10

(iii)

The regression equation shows that the number of errors reduce as the number of audits increases – this is demonstrated by the negative slope of the regression equation and confirmed by the correlation coefficient. This show that every time the independent variable x rises by one unit, the dependent variable y falls by 1.37 units, that is, as the number of audits increase the number of errors reduce. A graph of the variables will show this obvious relationship. 6 Marks

[Total: 20 Marks]

SOLUTION 5. (i) The beneficiary will receive €20,000 per year for 6 years. It is necessary to find the present value (PV) of these amounts. PV

=

20,000 x 1/(1 + 0.05)1 + 20,000 x 1/(1 + 0.05)2 + 20,000 x 1/(1 + 0.05)3 + 20,000 x 1/(1 + 0.05)4 + 20,000 x 1/(1 + 0.05)5 + 20,000 x 1/(1 + 0.05)6

This can be calculated for each year or use geometric progression formula where PV of annuity = Sn = a x (1 – rn)/(1 – r) and a = 20,000/ 1.05, r = 1/1.05 Therefore,

(ii)

(iii)

PV

=

20,000/1.05 x 1 – (1/1.05)6 / 1 – (1/1.05)

=

19,047.6 x (1 – 0.677)/(1 – 0.952)

=

19,047.6 x 6.729

=

€128,171

8 Marks

The interest will be compounded over the three years. The compound interest is derived as follows CI

=

P(1 + i)t - P where I = 20%, P = €15,000, t = 3.

CI

=

15,000(1 + 0.2)3 - 15,000

=

15,000 x 1.728 - 15,000

=

25,920 - 15,000 = €10,920

6 Marks

To find the probability Prob (x > 45,000), z = x1 - µ = 45,000 - 42,000 = 0.75 σ 4,000 From the normal tables, z = 0.75 gives 0.7734, that is, the area in the right hand tail is 1 – 0.7734 = 0.2266. Nearly 23% of tyres will last for more than 45,000 miles.

6 Marks

[Total: 20 Marks]

11

SOLUTION 6.

An explanation of the terms is provided below.

Statistical Hypotheses. A statistical hypothesis is an assertion about an attribute of a population which may also concern the type or nature of a population. To develop procedures for testing statistical hypotheses it is essential to know exactly what to expect when an hypothesis is true and we often hypothesize the opposite of what we hope to prove. For example if we wish to show that one method of teaching is more effective than another, we would hypotesize that the two methods are equally effective. Since we hypothesize that there is no difference in the two teaching methods, these hypotheses are called ‘null hypotheses’ and are denoted by Ho. The term ‘null hypothesis’ is used for any hypothesis that is set up primarily to see whether it can be rejected. The hypothesis that is used as an alternative to the null hypothesis, that is, the hypothesis that is accepted when the null hypothesis is rejected is the ‘alternative hypothesis’ and is denoted by H1. It ust always be formulated with the null hypothesis otherwise we would not know when to reject Ho. Alternative hypotheses usually specify that the population mean (or whatever other attribute may be of interest) is less than, greater than, or not equal to the value assumed under the null hypothesis. For any given problem, the choice of one of these alternatives depends on what we hope to be able to show, or where we want to put the burden of proof. 5 Marks

The Normal Distribution. Among many different continuous distributions used in statistics the most important is the normal distribution. It plays a very important role in the science of statistical inference since many business phenomena generate random variables with probability distributions that are well approximated by a normal distribution. The graph of a normal distribution is a bell-shaped curve that extends indefinitely in both directions. An important feature of normal distributions is that they depend only on two quantities µ and σ, that is, the mean and standard deviation. There are different curves depending on the value of µ and σ In the majority of work undertaken in this area the concern is with the ‘standard normal distribution’. The standard normal distribution is a normal distribution with µ = 0 and σ = 1. Areas under any normal curve are obtained by performing a change of scale that converts the units of measurement from the original scale, or x-scale, into standard units or ‘z-scores’ by means of the formula z = x –µ / σ In this new scale a value of z states how many standard deviations the corresponding value of x lies above or below the mean of its distribution. These values are obtained from standard normal tables to enable calculations to be easily performed. 5 Marks Internal Rate of Return. The internal rate of return (IRR) or the Yield is an alternative method of investment appraisal to Net Present Value. It can be described as the rate that a project earns. The decision rule when using IRR is that a firm should undertake a project if the annualised return in the form of the IRR is greater that the annual cost of capital (the rate of interest). If the IRR is less than the interest rate used to calculate NPV then the project would not be undertaken. In other words, if the IRR is less than the annual cost of capital then the company should avoid such capital expenditure. There is no precise formula for calculating the IRR of a given project, However, it can be estimated either graphically or by formula. Both of these techniques need the NPV calculated using two different discount rates. An advantage of using the IRR is that it does not depend on any external rates of interest. A disadvantage however is that it returns a relative (percentage) value and does not differentiate between the scale of projects, that is, one project could involve substantially larger cash flows than another. This could be of significance for some project comparisons. 5 Marks

The Appropriate Selection of a Base Period in Index Numbers. An index number measures the percentage change in value of an item relative to its value at a predetermined historical point known as the ‘base period’. At this base period the index number is equal to 100 and all subsequent index numbers are calculated as percentages of that base period. In a business application the choice of base period is very important. The base year should be a ‘normal’ representative period when no abnormal changes have occurred. If the period is abnormal, then future periods of relatively minor changes in price or quantity will be very difficult to reflex in the indes number; then then cover up any significant change in the value of the business variable being considered. It would be in appropriate to measure a variable such as the change in house prices if the base year selected was in the late 1980s as the boom in house prices over the past decade represented a time of boom for this sector of the economy. When making comparisons over long periods of time index numbers can become too large, too small or too similar to be meaningful. It is appropriate therefore, at times to reset the base period. If a comparison of index numbers is required over a period of time where there has been a change of base, appropriate calculations are made to reflect the change. 5 Marks 12